Question: Divide the following complex numbers. $ \dfrac{-12+26i}{-5+4i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-5-4i}$ $ \dfrac{-12+26i}{-5+4i} = \dfrac{-12+26i}{-5+4i} \cdot \dfrac{{-5-4i}}{{-5-4i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-12+26i) \cdot (-5-4i)} {(-5+4i) \cdot (-5-4i)} = \dfrac{(-12+26i) \cdot (-5-4i)} {(-5)^2 - (4i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-12+26i) \cdot (-5-4i)} {(-5)^2 - (4i)^2} = $ $ \dfrac{(-12+26i) \cdot (-5-4i)} {25 + 16} = $ $ \dfrac{(-12+26i) \cdot (-5-4i)} {41} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-12+26i}) \cdot ({-5-4i})} {41} = $ $ \dfrac{{-12} \cdot {(-5)} + {26} \cdot {(-5) i} + {-12} \cdot {-4 i} + {26} \cdot {-4 i^2}} {41} $ Evaluate each product of two numbers. $ \dfrac{60 - 130i + 48i - 104 i^2} {41} $ Finally, simplify the fraction. $ \dfrac{60 - 130i + 48i + 104} {41} = \dfrac{164 - 82i} {41} = 4-2i $